Error with "too many output arguments" using function "gradient"

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allen leung
allen leung 2018-8-9
评论： Allen Leung 2018-8-11
% %Step 1 Simulate true data
n=1e5;
mu_L=5000;%mm
sigma_L=100;%mm
mu_E=210e3;%N/mm^2
sigma_E=2000;%N/mm^2
mu_I=17500e4;%mm^4
sigma_I=10000;%mm^4
mu_A=136e2;%mm^2
sigma_A=1500;%mm^2
mu_fy=275;%N/mm^2
sigma_fy=30;%N/mm^2
%load
mu_load=2e6;
sigma_load=2e4;
%Simulate the true data
ui=rand(11,n);
uL=ui(1,:);%geometrical prop for experimental resistance data
uE=ui(2,:);%material prop for experimental resistance data
uI=ui(3,:);%geometrical prop for experimental resistance data
uA=ui(4,:);%geometricalprop for experimental resistance data
ufy=ui(5,:);%material prop for experimental resistance data
utL=ui(6,:);%geometrical prop for theoretical resistance data
utE=ui(7,:);%material prop for theoretical resistance data
utI=ui(8,:);%geometrical prop for theoretical resistance data
utA=ui(9,:);%geometricalprop for theoretical resistance data
utfy=ui(10,:);%material prop for theoretical resistance data
uload=ui(11,:);
L_e=mu_L-sqrt(2)*erfcinv(2*uL)*sigma_L;
E_e=mu_E-sqrt(2)*erfcinv(2*uE)*sigma_E;
I_e=mu_I-sqrt(2)*erfcinv(2*uI)*sigma_I;
A_e=mu_A-sqrt(2)*erfcinv(2*uA)*sigma_A;
fy_e=mu_fy-sqrt(2)*erfcinv(2*ufy)*sigma_fy;
L_t=mu_L-sqrt(2)*erfcinv(2*utL)*sigma_L;
E_t=mu_E-sqrt(2)*erfcinv(2*utE)*sigma_E;
I_t=mu_I-sqrt(2)*erfcinv(2*utI)*sigma_I;
A_t=mu_A-sqrt(2)*erfcinv(2*utA)*sigma_A;
fy_t=mu_fy-sqrt(2)*erfcinv(2*utfy)*sigma_fy;
alpha=0.34; %buckling curve b, buckling along y-y axis
Ncr_e=((pi())^2)*E_e.*I_e./(L_e.^2);
lamda_bar_e=sqrt(A_e.*fy_e./Ncr_e);
phi_e=0.5*(1+alpha*(lamda_bar_e-0.2)+lamda_bar_e.^2);
chi_e=1./(phi_e+sqrt(phi_e.^2-lamda_bar_e.^2));
Nb_Rd_e=chi_e.*A_e.*fy_e;
alpha=0.34; %buckling curve b, buckling along y-y axis
Ncr_t=((pi())^2)*E_t.*I_t./(L_t.^2);
lamda_bar_t=sqrt(A_t.*fy_t./Ncr_t);
phi_t=0.5*(1+alpha*(lamda_bar_t-0.2)+lamda_bar_t.^2);
chi_t=1./(phi_t+sqrt(phi_t.^2-lamda_bar_t.^2));
Nb_Rd_t=chi_t.*A_t.*fy_t;
b=sum(Nb_Rd_e.*Nb_Rd_t)/sum(Nb_Rd_t.^2);
plot(Nb_Rd_t,Nb_Rd_e)
%The true b value is used to generate samples
%calculate the characteristic resistance according to eurocode Annex D
error=Nb_Rd_e./(b*Nb_Rd_t);
%Take natural log
delta=log(error);
%estimated value of expected value of error
delta_bar=sum(delta)/n;
%estimated error variance
error_var=(sum(delta-delta_bar)^2)/(n-1);
%coefficient of variance of error
v_delta=sqrt(exp(error_var^2)-1);
%Assessment via the characteristic value
Xd_L=mu_L*(1-1.64*sigma_L/mu_L);%kn chosen as 1.64 due to large no. of tests
Xd_E=mu_E*(1-1.64*sigma_E/mu_E);
Xd_I=mu_I*(1-1.64*sigma_I/mu_I);
Xd_A=mu_A*(1-1.64*sigma_A/mu_A);
Xd_fy=mu_fy*(1-1.64*sigma_fy/mu_fy);
alpha=0.34; %buckling curve b, buckling along y-y axis
Ncr_t=((pi())^2)*E_t.*I_t./(L_t.^2);
lamda_bar_t=sqrt(A_t.*fy_t./Ncr_t);
phi_t=0.5*(1+alpha*(lamda_bar_t-0.2)+lamda_bar_t.^2);
chi_t=1./(phi_t+sqrt(phi_t.^2-lamda_bar_t.^2));
Nb_Rd_t=chi_t.*A_t.*fy_t;
%summation of variance
%partial derivative of L
[pL,pE,pI,pA,pfy]=gradient(Nb_Rd_t);
sum_var=((pL*sigma_L).^2)+((pE*sigma_E).^2)+((pI*sigma_I).^2)+((pA*sigma_A).^2)+((pfy*sigma_fy).^2);
%resistance evaluated using mean values of basic variables
alpha=0.34; %buckling curve b, buckling along y-y axis
Ncr_m=((pi())^2)*mu_E.*mu_I./(mu_L.^2);
lamda_bar_m=sqrt(mu_A.*mu_fy./Ncr_m);
phi_m=0.5*(1+alpha*(lamda_bar_m-0.2)+lamda_bar_m.^2);
chi_m=1./(phi_m+sqrt(phi_m.^2-lamda_bar_m.^2));
Nb_Rd_m=chi_m.*mu_A.*mu_fy;
%coefficient of variation Vrt
v_rt_square=sum_var/(Nb_Rd_m^2);
%coefficient of variation Vr_square
vr_square=v_delta^2+v_rt_square;
%Q
Q=sqrt(log(vr_square+1));
%rk
rk=b*Nb_Rd_m*exp(-1.64*Q-0.5*(Q^2))
end end
I want to evaluate the partial derivative of Nb_Rd_t with respect to five different variables L_t, E_t, I_t, A_t, fy_t. These variables are 1 by 1e5 vectors. I do not understand why there are too many output arguments. Thanks for your answers in advance!!
  3 个评论
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Jan
Jan 2018-8-9
Prefer to post code as text using the "{} Code" button. This is usually more convenient than a screenshot.
allen leung
allen leung 2018-8-9
Sorry for causing you convenience guys! I have just inserted the code and attached my m file here. Thanks!

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采纳的回答

Jan
Jan 2018-8-9
gradient is a built-in function, which has as many outputs as the input has dimensions. See:
whos gradient -all
Use a different name for your function, e.g. "myGradient".
  5 个评论
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Jan
Jan 2018-8-11
To which 5 variables? You provide a [1 x 10000] vector currently. I cannot know, what you want to do instead.
Allen Leung
Allen Leung 2018-8-11
These 5 variables L_t, E_t, I_t, A_t and fy_t contribute to the value of Nb_Rd_t through these equations. All the variables are 1x10000 vectors. I understand gradient doesn't work in this case. But may I know if there is other solution to find out the partial derivatives corresponding to these variables?
if true
% alpha=0.34; %buckling curve b, buckling along y-y axis
Ncr_t=((pi())^2)*E_t.*I_t./(L_t.^2);
lamda_bar_t=sqrt(A_t.*fy_t./Ncr_t);
phi_t=0.5*(1+alpha*(lamda_bar_t-0.2)+lamda_bar_t.^2);
chi_t=1./(phi_t+sqrt(phi_t.^2-lamda_bar_t.^2));
Nb_Rd_t=chi_t.*A_t.*fy_t;
end

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